Difference between revisions of "Falling Body with Air Resist"
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v(t) = (mg/k)(1-e<sup>-kt/m</sup>) | v(t) = (mg/k)(1-e<sup>-kt/m</sup>) | ||
+ | |||
+ | when t goes to infinity, e<sup>-kt/m</sup> goes to 0 so the terminal velocity = mg/k |
Revision as of 07:06, 5 May 2020
Ftotal = Fgrav - Fair
ma = mg - kv where k units are kg/sec
mdv/dt = mg - kv
put into standard order:
mdv/dt + kv = mg
dv/dt + vk/m = g
find u = e∫k/mdt = ekt/m
multiply by u:
ekt/mdv/dt + ekt/mvk/m = gekt/m
d/dt(vekt/m) = gekt/m
integrate both sides:
vekt/m + C1 = gm/kekt/m + C2
solve for v:
v = mg/k - (C1 + C2)/ekt/m
v = mg/k - Ce-kt/m
If at time 0, the body is at rest then v(0) = 0 = mg/k - C so C = mg/k
therefore v = mg/k - mg/ke-kt/m
or
v(t) = (mg/k)(1-e-kt/m)
when t goes to infinity, e-kt/m goes to 0 so the terminal velocity = mg/k